Prove that the largest isosceles triangle that can be drawn in a circle, is an equilateral triangle?

1 Answer
Mar 19, 2017

Please see below.

Explanation:

Let their be an isosceles triangle ABC inscribed in a circle as shown, in which equal sides AC and BC subtend an angle x at the center. It is apparent that side AB subtends an angle 360^0-x at the center (as shown). Note that for equilateral triangles all these angles will be (2pi)/3.

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As the area of the triangle portion subtended by an angle x is R^2/2sinx,

the complete area of triangle ABC is

A=R^2/2(sinx+sinx+sin(360-2x)

= R^2/2(2sinx-sin2x)

= R^2(sinx-sinxcosx)

= R^2sinx(1-cosx)

For maximization we should have (dA)/(dx)=0

i.e. R^2(cosx(1-cosx)+sinx xx sinx)=0

or cosx-cos^2x+1-cos^2x=0

or 2cos^2x-cosx-1=0

or 2cos^2x-2cosx+cosx-1=0

or 2cosx(cosx-1)+1(cosx-1)=0

or (2cosx+1)(cosx-1)=0

Hence cosx=-1/2 or cosx=1

i.e. x=(2pi)/3 or x=0

But for a triangle x!=0

hence x=(2pi)/3

and hence for maximum area triangle must be equilateral.